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A nucleus X initially at rest, undergoes alpha decay according to the equation `_Z^232Xrarr_90^AY+alpha` What fraction of the total energy released in the decay will be the kinetic energy of the alpha particle?A. (a) `90/92`B. (b) `228/232`C. (c) `sqrt(228/232)`D. (d) `1/2` |
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Answer» Correct Answer - B `A=232-4=228` From conservation of momentum, `p_alpha=p_gamma=sqrt(2K_alpham_alpha)=sqrt(2K_gammam_gamma)` or `(K_alpha)/(K_gamma)=(m_gamma)/(m_alpha)=228/4` `:.` `K_alpha=((228)/(228+4))K_(t otal)` `=(228/232)K_(T otal)` |
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